finding the cosine

[homeschool girl]

The problem:

In triangle [MATH]ABC[/MATH], [MATH]AB = AC,[/MATH] and angle [MATH]A[/MATH] is equal to [MATH]36^\circ.[/MATH] Point [MATH]D[/MATH] is on [MATH]\overline{AC}[/MATH] so that [MATH]\overline{BD}[/MATH] bisects [MATH]\angle ABC.[/MATH]

(a) Prove that [MATH]BC = BD = AD[/MATH].


(b) Let [MATH]x = BC[/MATH] and let [MATH]y = CD[/MATH]. Using similar triangles [MATH]BCD[/MATH] and [MATH]ABC[/MATH], write an equation involving [MATH]x[/MATH] and [MATH]y[/MATH].


(c) Let [MATH]r = \frac{y}{x}[/MATH]. Write the equation from Part (b) in terms of [MATH]r,[/MATH] and find [MATH]r.[/MATH]

(d) Find [MATH]\cos 36^\circ[/MATH] and [MATH]\cos 72^\circ[/MATH] using Parts a to c. (Do not use your calculator!)


here's what I have so far:

Part A

[MATH]\triangle ABC[/MATH]is icocilese, so [MATH]\angle ABC = \angle ACB = \frac{180-36}{2}=72[/MATH].


adding the angle bisector [MATH]BD[/MATH], [MATH]\angle DBC = \frac{72}{2} = 36[/MATH].


So [MATH]\triangle ABC \sim \triangle BCD[/MATH]with AA similararity.


We can find the side lengths of the triangles relative to each other, where [MATH]x[/MATH] is the length of BC and [MATH]r[/MATH] is the ratio of [MATH]\triangle BCD[/MATH] to [MATH]\triangle ABC[/MATH].

using the angle bisector theorem, we can solve for the ratio.


[MATH]\frac{x\div r}{x\div r - x \cdot r} = \frac{x}{x \cdot r}[/MATH]

[MATH]\frac{1}{1 - r^2} = \frac{1}{r}[/MATH]

[MATH]1- r^2 -r = 0[/MATH]

[MATH]r=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(-1)}}{2\cdot1}[/MATH]

[MATH]r=\frac{-1\pm\sqrt{5}}{2}[/MATH]

because [MATH]D[/MATH]is on [MATH]\overline AC[/MATH], [MATH]x\cdot r < x\div r, so -1>r>1[/MATH], meaning [MATH]r\neq\frac{-1-\sqrt{5}}{2}[/MATH]because [MATH]\frac{-1-\sqrt{5}}{2} < -1[/MATH].


plugging in the value of [MATH]r[/MATH]into AD, we get

[MATH]AD = x\div \frac{-1+\sqrt{5}}{2} -x \cdot \frac{-1+\sqrt{5}}{2}[/MATH]
[MATH]AD = \frac{x\cdot(\sqrt{5}+1)}{2} - \frac{-x+\sqrt{5}x}{2}[/MATH]
[MATH]AD = \frac{\sqrt{5}x+x+x-\sqrt{5}x}{2}[/MATH]
[MATH]AD = \frac{2x}{2}[/MATH]
[MATH]AD = x[/MATH]

Which means [MATH]\boxed{BC = BD = AD}[/MATH].


Part b


We set [MATH]y[/MATH]equal to [MATH]x\cdot r[/MATH], our old value of [MATH]CD[/MATH]. This sets [MATH]r[/MATH]equal to [MATH]\frac{y}{x}[/MATH]. We can define [MATH]AB[/MATH]and [MATH]AC[/MATH]both as [MATH]x \div \frac{y}{x} = \frac{x^2}{y}[/MATH], or as [MATH]x+y[/MATH].

So, [MATH]\boxed{\frac{x^2}{y}=x+y}[/MATH].


Part c


We write the equation from above but using [MATH]r[/MATH] instead of [MATH]\frac{y}{x}.[/MATH]
[MATH]x \div r = x+y.[/MATH]
[MATH]x = (x+y)\cdot r.[/MATH]
[MATH]\frac{x}{x+y}=r.[/MATH]
[MATH]\boxed{r=\frac{x}{x+y}}.[/MATH]


I'm stuck on how to do part d.

I tried dropping an altitude from C to AD but I don't really know where to go from there

 

[Dr.Peterson]

Part a can be done more quickly, just finding isosceles triangles. Most of the work you did there looks like the work for b and c and beyond.

Since you've already found the value of r, use it with the law of cosines and/or the law of sines with one or another of the triangles in the picture to answer d. Or, you could drop a perpendicular from A to BC, and from D to AB.